# The mathematical denotion of why each bidder bids the expected 2nd highest bidder's value in first-price auctions

Advanced Microeconomics by Jehle and Reny (p431) contains the following theorem:

Because FN-1(·) is the distribution function of the highest value among a bidder’s N - 1 competitors, the bidding strategy displayed in Theorem 9.1 says that each bidder bids the expectation of the second highest bidder’s value conditional on his own value being highest. But, because the bidders use the same strictly increasing bidding function, having the highest value is equivalent to having the highest bid and so equivalent to winning the auction. So, we may say: In the unique symmetric equilibrium of a first-price, sealed-bid auction, each bidder bids the expectation of the second-highest bidder’s value conditional on winning the auction.

I don't understand the maths of this. I do sense it's logical you'd only bid what you expect to be the value the 2nd highest bidder has for the object, but I can't see how the equation of theorom 9.1 leads you to that conclusion. Could someone explain that to me? Please consider I'm not as proficient with mathematics as many of you are (otherwise I probably wouldn't need to make this thread), so I need to have this explained step-by-step, if you don't mind.

• For me it's the other way around. I can see how the formula says that you'd bid the expectation of the second-highest bidder's value conditional on winning the auction, but I don't sense why this should be logical. Mar 6, 2023 at 8:37
• Could you then explain how the formula shows how you'd bid the expectation of the second-highest bidder's value, conditional on winning the auction? That is my purpose of this thread. Mar 8, 2023 at 10:51

First, assume $$N-1$$ bidders' values $$v_i$$ are independently drawn from some interval, let's say $$[0,1]$$, according to the cdf $$F$$. Now fix some $$x$$ in this interval. What's the probability $$P(m\le x)$$ that the maximum $$m=\max_iv_i$$ is at most $$x$$?
The maximum of the $$v_i$$s is at most $$x$$ iff all the single $$v_i$$s are at most $$x$$, i.e. iff $$v_1\le x$$ and $$v_2\le x$$ and ... $$v_{N-1}\le x$$. For each $$i$$ we have that the probability that $$v_i\le x$$ is $$P(v_i\le x)=F(x)$$ by definition of $$F$$. By independence these probabilities have to be multiplied to yield the total probability. Thus the probability that the maximum of the $$v_i$$s is at most $$x$$ is $$P(m\le x)=F^{N-1}(x)$$. Therefore $$F^{N-1}$$ is the distribution function of this maximum.
Now assume your own value is $$v$$. What's the probability $$P(win)$$ that you win the auction? For this, the maximum value of the other $$N-1$$ bidders has to be below $$v$$. The probability of this is $$F^{N-1}(v)$$, as we just argued, thus $$P(win)=F^{N-1}(v)$$.
Now let $$x\in[0,1]$$. What's the probability $$Q$$ that (i) you win AND (ii) $$m\le x$$? These two events are not independent, but we know that it can be written in two ways using conditional probabilities, either as $$Q=P(win)P(m\le x|win)$$ or as $$Q=P(m\le x)P(win|m\le x)$$. Rearranging gives us Bayes' Theorem for this case, $$P(m\le x|win)=P(m\le x)P(win|m\le x)/P(win)$$.
Assume now that $$x. Then of course (ii) already implies (i), so $$P(win|m\le x)=1$$ and the equation simplifies to $$P(m\le x|win)=P(m\le x)/P(win)$$. Substituting the values we calculated above, $$P(m\le x|win)=\frac{1}{F^{N-1}(v)}F^{N-1}(x)$$.
The lhs of this equation is the distribution function for the second highest bidder's valuation, conditional on you winning the auction. What we are looking for is the expectation of this. To find this, we integrate over all possible values of $$x$$, i.e. over $$[0,v]$$, the product of $$x$$ and the differential of the distribution function. Doing this on the rhs of the equation and pulling the factor $$\frac{1}{F^{N-1}(v)}$$ out of the integral finally gives us the formula in the book.